Show how would you synthesise the following alcohols from appropriate alkenes? (i) Pentan-2-ol (ii) 2-Methylbutan-2-ol (iii) Pentan-1-ol
Pentan-2-ol can be synthesised by acid-catalysed hydration of pent-2-ene. The reaction proceeds via Markovnikov addition where the OH group attaches to the more substituted carbon.
CH₃—CH=CH—CH₂—CH₃ + H₂O (dil. H₂SO₄) → CH₃—CH(OH)—CH₂—CH₂—CH₃ (Pentan-2-ol)
2-Methylbutan-2-ol can be synthesised by acid-catalysed hydration of 2-methylbut-2-ene. The reaction follows Markovnikov's rule giving the tertiary alcohol.
CH₃—CH₂—C(CH₃)=CH₂ + H₂O (dil. H₂SO₄) → CH₃—CH₂—C(CH₃)(OH)—CH₃ (2-Methylbutan-2-ol)
Pentan-1-ol can be synthesised from pent-1-ene using hydroboration-oxidation. This method gives anti-Markovnikov addition where the OH group attaches to the less substituted carbon, producing the primary alcohol.
CH₃—CH₂—CH₂—CH=CH₂ → (BH₃, then H₂O₂/NaOH) → CH₃—CH₂—CH₂—CH₂—CH₂OH (Pentan-1-ol)
Explanation
The context shows that hydration of propene in presence of dilute sulphuric acid produces propan-2-ol (secondary alcohol), demonstrating acid-catalysed hydration follows Markovnikov's rule. For secondary and tertiary alcohols (parts i and ii), acid-catalysed hydration of appropriate alkenes gives the desired products. For primary alcohols like pentan-1-ol, acid-catalysed hydration would give the Markovnikov product (pentan-2-ol), so hydroboration-oxidation is required for anti-Markovnikov addition.